Let $f$ be a vector-valued function defined by $f(t)=(\log_4(t),t^6-9t^3)$. Find $f$ 's second derivative $f''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\dfrac{1}{t^2\ln(4)},30t^4-54t\right)$ (Choice B) B $\left(\dfrac{1}{t\ln(4)},6t^5-27t^2\right)$ (Choice C) C $\left(\dfrac{1}{t^2},6t^4-27t\right)$ (Choice D) D $\left(-\dfrac{1}{t^2},-30t^4-54t\right)$
Answer: We are asked to find the second derivative of $f$. This means we need to differentiate $f$ twice. In other words, we differentiate $f$ once to find $f'$, and then differentiate $f'$ (which is a vector-valued function as well) to find $f''$. Recall that $f(t)=(\log_4(t),t^6-9t^3)$. Therefore, $f'(t)=\left(\dfrac{1}{t\ln(4)},6t^5-27t^2\right)$. Now let's differentiate $f'(t)=\left(\dfrac{1}{t\ln(4)},6t^5-27t^2\right)$ to find $f''$. $f''(t)=\left(-\dfrac{1}{t^2\ln(4)},30t^4-54t\right)$ In conclusion, $f''(t)=\left(-\dfrac{1}{t^2\ln(4)},30t^4-54t\right)$.